# Mechanics - two boxes down a hill

1. Sep 25, 2011

### Hioj

1. The problem statement, all variables and given/known data
The first box slides down a hill and gains speed and ends at $v=1 \dfrac{m}{s}.$

The second box has an initial speed of $v_{0}=1 \dfrac{m}{s}$ and slides down the same hill. What speed does it end up with?

The boxes have equal masses.

Solve this without first solving for the height $h$. This is intuition practice.

2. Relevant equations
$\dfrac{1}{2}mv^{2}=mgh$

3. The attempt at a solution
Since $\dfrac{1}{2}mv^{2}=mgh$, then $v=\sqrt{mgh}$. The first gains a speed of $1 \dfrac{m}{s}$, so the second must also gain $1 \dfrac{m}{s}$. So it ends up at $2 \dfrac{m}{s}$?

I remember the answer being $\sqrt{2}$.

Last edited: Sep 25, 2011
2. Sep 25, 2011

### Spinnor

You have,

v = (m*g*h)^.5

It should be v = (2*g*h)^.5

Why don't you solve for h just for fun?

The second box will have twice the energy of the first but as kinetic energy goes as v^2 the velocity won't be double.

3. Sep 26, 2011

### Hioj

The following assumes $g=10\dfrac{m}{s}$ for simplicity.

We have
$$\dfrac{1}{2}mv^{2}=mgh$$
and so
$$h=\dfrac{v^{2}}{2g}=\dfrac{1}{20}$$ as $v^{2}=1\dfrac{m}{s}$ for the first box.
For the second box, it must be true that the kinetic energy with the end speed $v$ equals the kinetic energy it starts off with $(\dfrac{1}{2}mv_{0}^{2})$ plus the gained potential energy. Then
$$\dfrac{1}{2}mv^{2}=\dfrac{1}{2}mv_{0}^{2}+mgh$$
$$\Longrightarrow v^{2}=v_{0}^{2}+2gh=1+\dfrac{2g}{20}=1+1=2$$
$$\Longrightarrow v=\sqrt{2}.$$

Correct?

Could this be done any easier?